TITLE.PM5

166 ENGINEERING THERMODYNAMICS

dharm

/M-therm/Th4-4.pm5

Q = 55 kJ/s = 55 kJ / s 10

60 kg/s

= 55 × 6 = 330 kJ/kg

Now, ∆KE = CC^2

2

1

(^222)
2
190 110
2
− =()()− Nm or J = 12000 J or 12 kJ/kg
∆PE = (Z 2 – Z 1 ) g = (55 – 0) × 9.81 Nm or J = 539.5 J or ≈ 0.54 kJ/kg
Substituting the value in steady flow equation,

• 330 = 12 + 0.54 – 105.77 + W or W = – 236.77 kJ/kg.

Work done per second = – 236.77 ×

10
60 = – 39.46 kJ/s = – 39.46 kW. (Ans.)
Example 4.36. In a gas turbine unit, the gases flow through the turbine is 15 kg/s and the
power developed by the turbine is 12000 kW. The enthalpies of gases at the inlet and outlet are
1260 kJ/kg and 400 kJ/kg respectively, and the velocity of gases at the inlet and outlet are
50 m/s and 110 m/s respectively. Calculate :
(i)The rate at which heat is rejected to the turbine, and
(ii)The area of the inlet pipe given that the specific volume of the gases at the inlet is
0.45 m^3 /kg.
Solution. Rate of flow of gases, m& = 15 kg/s
Volume of gases at the inlet, v = 0.45 m^3 /kg
Power developed by the turbine, P = 12000 kW

∴ Work done, W =^1200015 = 800 kJ/kg

Enthalpy of gases at the inlet, h 1 = 1260 kJ/kg

Enthalpy of gases at the oulet, h 2 = 400 kJ/kg

Velocity of gases at the inlet, C 1 = 50 m/s

Velocity of gases at the outlet, C 2 = 110 m/s.

Fig. 4.45